I have two questions:

1) When one says an ARMA process is 'stationary,' do they mean strongly stationary or weakly stationary?

2) Is there a quick way to find the variance of a stationary AR(2) model $$y_t = \beta_1 y_{t-1} + \beta_2 y_{t-2} + \epsilon_t?$$ The only way I can think of doing this is by multiplying by $y_t$, $y_{t-1}$ and $y_{t-2}$, taking expectations, and solving the Yule-Walker system with 3 equations and 3 unknowns. The trick for AR(1) models, where one takes expectations of both sides, doesn't slide here because you get a $\mathrm{Cov}(y_{t-1}, y_{t-2})$ term.

  • 1
    $\begingroup$ Regarding (1), it must depend on the context. An answer claiming otherwise would be careless, IMHO. $\endgroup$ Commented Jan 19, 2017 at 17:39
  • $\begingroup$ Regarding (2), the variance of the AR(2) model will depend on the first and second order autocovariances. As you say, in order to get its value you will have to solve the Yule-Walker equations; by doing so, I think you will get $Var(y_t)=\frac{(1-\beta_2)\sigma^2_\varepsilon}{(1+\beta_2)[(1-\beta_2)^2-\beta_1^2]}$. $\endgroup$
    – javlacalle
    Commented Jan 19, 2017 at 18:10
  • $\begingroup$ +1 @RichardHardy Please allow me to add a plug for precisely communicating one's methods, and results! $\endgroup$
    – Alexis
    Commented Jan 15, 2020 at 18:36

4 Answers 4



Considering your AR(2) process with mean zero i.i.d. noise $\epsilon_t$ of variance $\sigma_\epsilon^2$,

$$ y_t = \beta_1 y_{t-1} + \beta_2 y_{t-2} + \epsilon_t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(*) $$

we can rewrite it in terms of the lag operator $L$

$$ (1 - \beta_1 L - \beta_2 L^2)y_t = \epsilon_t $$

so that we have a new operator

$$ 1 - \beta_1 L - \beta_2 L^2 $$

If the above were a polynomial in $L$, let its roots be $z_1^{-1}, z_2^{-1}$. We call the polynomial the characteristic polynomial, and $z_1, z_2$ its factors. (Note the factors and roots are inverses of each other, and may be complex.)

It can be shown that the AR(2) is stationary when $|z_1| < 1$ and $|z_2|<1$. i.e. when all of the following are met: $$ |\beta_2| < 1 \\ \beta_2 + \beta_1 < 1 \\ \beta_2 - \beta_1 < 1 $$

For details, see this answer. Or in terms of the more general ARMA(p,q) process, see Introduction to Time Series and Forecasting. Brockwell and Davis. 2016. p 74.


If the process is stationary, we can write the covariance as a function of increment alone, so let the covariance function $\gamma(k) \doteq E[y_t y_{t+k}]$. We can find the variance $\gamma(0)$ by squaring and taking the expectation of both sides of equation $(*)$ with the following result

$$ \gamma(0) = \beta_1^2 \gamma(0) + \beta_2^2 \gamma(0) + 2 \beta_1 \beta_2 \gamma(1) + \sigma_\epsilon^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(**) $$

Starting again from equation $(*)$, this time multiply both sides by $y_{t-k}$ and again take the expectation

$$ \gamma(k) = \beta_1 \gamma(k-1) + \beta_2 \gamma(k-2) $$

Usefully, we now have an expression for covariance, but we just need it to compute $\gamma(1)$. Since $\gamma(-1) = \gamma(1)$ (examine the definition, note that covariance is not a function of $t$), we can let $k=1$ in the above equation and get

$$ \gamma(1) = \frac{\beta_1 \gamma(0)}{1-\beta_2} $$

Substituting into $(**)$, we get the variance

$$ \text{Var}(y_t) = \gamma(0) = \frac{(1-\beta_2)\sigma_\epsilon^2}{(1+\beta_2)(1 - \beta_1 - \beta_2)(1 + \beta_1 - \beta_2)} $$

As a sanity check, our stationarity conditions on $\beta_1, \beta_2$ described earlier are precisely the conditions which make our expression for $\text{Var}(y_t)$ positive.

The above is essentially a summary and rearrangement of some parts of these notes.

Strong or Weak Stationarity

I have only seen arguments of AR process stationarity in terms of fixed mean and variance, so weak stationarity is implied. However, if the stationary distribution can be characterized completely in terms of those first and second moments, then we also have strong stationarity. See A unified view of linear AR(1) models. G.K. Grunwald. 1996.

As an example, if we have an AR(1) process and $\epsilon_t$ is i.i.d. Gaussian, then the process' stationary distribution is also Gaussian. Since the Gaussian is fully specified by its first two moments, we have strong stationarity in that case. I am unsure whether or not this also applies to more general AR(p) Gaussian processes, or to AR(p) processes with other kinds of i.i.d. noise.

  • $\begingroup$ I have seen the notes you mention above. However I have not been able to find your last equation from (**) your $\gamma(1)$ Equation. Could you please help? Thakns $\endgroup$ Commented Feb 9, 2022 at 20:07
  • $\begingroup$ Can you include more detail @Herman? I don’t understand your question yet. $\endgroup$
    – kdbanman
    Commented Feb 10, 2022 at 15:57
  • $\begingroup$ Unfortunately there is no room here to show you my computations. I wish I can include a file on this comment. I cannot get your equation for Var(y_t) doing what you are saying. That is, plugging he expression for $\gamma(1)$ in Equation (**). Thanks $\endgroup$ Commented Feb 11, 2022 at 13:14
  • $\begingroup$ I did it my own way. I will post my computations. $\endgroup$ Commented Feb 11, 2022 at 17:07

I do not have enough reputation to comment, so I’ll post it as an answer instead:

The accepted answer can be simplified to:

$$ \sigma = \frac{\sigma_Z^2}{1-\beta_1^2 (\frac{1+\beta_2}{1-\beta_2})-\beta_2^2} $$

See also:

Prove the Variance of an AR(2) Model

EDIT: So I re-did the math and it seems the answer of the posted link contains a small mistake. I updated the simplified answer accordingly.


Stack it, i.e. write as VAR(1) for the vector x(t) = [y(t),y(t-1)]', x(t) = [beta1,beta2;1,0] x(t-1) + [epsilon(t),0]'. Vectorize the resulting matrix equation and exploit vec(ABC)=(C' kron A) vec(B)

  • 4
    $\begingroup$ Welcome to CV, Harald Uhlig. Your answer would be made much more clear by editing it to use MathJax (a LaTeX-like mark-up language). You can get a feel for this by examining a CV question or answer by clicking it's "edit" button (not that you need to actually suggest an edit when doing so! :). You can similarly edit your own answer using the "edit" link at lower left. $\endgroup$
    – Alexis
    Commented Jan 15, 2020 at 18:40

From $X_t = \phi_1 X_{t-1} + \phi_2 X_{t-2} $,

\begin{eqnarray*} E[ X_t X_{t-k}] &=& \phi_1 E[X_{t-1} X_{t-k}] + \phi_2 E[X_{t-2} X_{t-k}] + E[X_t X_{t-1}] \\ \gamma_k &=& \phi_1 \gamma_{k-1} + \phi_2 \gamma_{k-2} + E[Z_t X_{t-k}] \quad (*) \end{eqnarray*} Now, \begin{eqnarray} E[Z_t X_{t-k}] = \left \{ \begin{array}{cc} \sigma_Z^2 & \text{if } k = 0 \\ 0 & \text{if } k=1,2, \cdots \end{array} \right . (**) \end{eqnarray}

Then combining (*) with (**) and using the symmetry property of the autocorrelation function. That is, $\gamma_t = \gamma_{-t}$

\begin{eqnarray} \sigma^2 = \gamma_0 &=& \phi_1 \gamma_{1} + \phi_2 \gamma_{2} + \sigma_Z^2 \quad , \quad (****) \\ \gamma_k &=& \phi_1 \gamma_{k-1} + \phi_2 \gamma_{k-2} \quad , \quad k=1,2, \cdots. \quad , \quad (***) \end{eqnarray}

We need to find $\gamma_1$ and $\gamma_2$. From () \begin{eqnarray} \gamma_1 = \phi_1 \gamma_0 + \phi_2 \gamma_{-1} = \phi_1 \gamma_0 + \phi_2 \gamma_1 \\ \gamma_1 = \frac{\phi_1 \gamma_0}{1 - \phi_2} = \frac{\phi_1 \sigma^2}{1 - \phi_2} \end{eqnarray} From () with $k=2$ we find that \begin{eqnarray} \gamma_2 = \phi_1 \gamma_1 + \phi_2 \gamma_0 = \phi_1 \frac{\phi_1 \sigma^2}{1-\phi_2} + \phi_2 \sigma^2 = \sigma^2 \left ( \frac{\phi_1^2 + \phi_2(1 - \phi_2) }{1 - \phi_2}\right ) \quad , \quad (*****) \end{eqnarray}

Then by pluggin this in (****) $$ \sigma^2 = \phi_1 \left ( \frac{\phi_1 \sigma^2}{1 - \phi_2} \right ) + \phi_2 \sigma^2 \left ( \frac{\phi_1^2 + \phi_2(1 - \phi_2) }{1 - \phi_2}\right ) + \sigma_Z^2$$

That is,

\begin{eqnarray} \sigma^2 \left [1 - \left ( \frac{\phi_1^2}{1 - \phi_2} \right ) -\phi_2 \left ( \frac{\phi_1^2 + \phi_2(1 - \phi_2) }{1 - \phi_2}\right ) \right ] = \sigma_Z^2 \end{eqnarray}

We simplify the quantity in brackets. That is:

\begin{eqnarray} 1 - \left ( \frac{\phi_1^2}{1 - \phi_2} \right ) -\phi_2 \left ( \frac{\phi_1^2 + \phi_2(1 - \phi_2) }{1 - \phi_2}\right ) &=&\frac{1 -\phi_2 - \phi_1^2 - \phi_2 \phi_1^2 - \phi_2^2 + \phi_2^3}{1 - \phi_2} \\ &=& \frac{(1 - \phi_2^2) - \phi_1^2(1 + \phi_2) + \phi_2(\phi_2^2 -1)}{1 - \phi_2} \\ &=& \frac{(1-\phi_2)(1 + \phi_2) - \phi_1^2(1 + \phi_2) - \phi_2(1 -\phi_2)(1 + \phi_2) }{1-\phi_2} \\ &=& \frac{(1 + \phi_2) (1 - \phi_2 - \phi_1^2 - \phi_2(1 - \phi_2))}{1 - \phi_2} \\ &=& \frac{(1 + \phi_2) (1 - 2 \phi_2 + \phi_2^2 - \phi_1^2)}{1 - \phi_2} \\ &=& \frac{(1 + \phi_2)[( 1 - \phi_2)^2 - \phi_1^2]}{1 - \phi_2} \\ &=& \frac{(1+ \phi_2)( 1 - \phi_2 - \phi_1)(1 - \phi_2 + \phi_1)}{1 - \phi_2} \end{eqnarray} We find then that \begin{eqnarray} \sigma^2 = \frac{(1 - \phi_2) \sigma_Z^2}{(1+ \phi_2)( 1 - \phi_2 - \phi_1) (1 - \phi_2 + \phi_1)} \end{eqnarray}


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